CODEWITHSUNDEEP
javalistcollectionsjava-7
AJJ
AJJ

Reputation: 3628

Sum List of BigDecimal

I am working on a project with Spring and Maven and Java 7.

I have a list with bigdecimal and have to sum up all the elements in the list. i know using for loop we can to as below,

List<BigDecimal> list = new ArrayList<BigDecimal>();
list.add(new BigDecimal(10.333));
list.add(new BigDecimal(14.333));
BigDecimal result = new BigDecimal(0);
for (BigDecimal b : list) {
     result = result.add(b);
}

Is there a better way to do? using google gauva FluentIterable or apache ArrayUtils/StatUtils?

Upvotes: 8

Views: 21870

Answers (3)

Daniel Beck
Daniel Beck

Reputation: 111

Here is a way to do it with only Java 8 and streams:

List<BigDecimal> list = Arrays.asList(new BigDecimal(10.333), //
   new BigDecimal(14.333));
BigDecimal result = list.stream().reduce(BigDecimal.ZERO, BigDecimal::add);

Upvotes: 10

Donald Raab
Donald Raab

Reputation: 6706

If you don't mind the extra third-party library dependency, then there is a solution for this problem in Eclipse Collections. 

List<BigDecimal> list =
    Lists.mutable.with(
        new BigDecimal("10.333"),
        new BigDecimal("14.333"));

Assert.assertEquals(
    new BigDecimal("24.666"),
    Iterate.sumOfBigDecimal(list, Functions.getPassThru()));

Since you have a list of BigDecimal already, the function that is applied to each just passes the value through.

Note: I am a committer for Eclipse Collections

Upvotes: 2

chrylis -cautiouslyoptimistic-
chrylis -cautiouslyoptimistic-

Reputation: 77206

If you can't use Java 8 lambdas or Groovy closures, in which case you could write a one-liner to replace your four lines, the code you have is thoroughly clear. Using a library-based iteration tool will almost certainly make the code more complicated and will certainly make it slower. You made it about as simple as it gets; it's fine.

Upvotes: 3

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